All Answers 353090 Answers

70+40=110 students with either condition

110-20 (because there are 20 repeats)=90 teachers with a condition

120-90=30 without conditions 

30/120=25%

Have you tried drawing the picture.  

This question really is not very hard.

Once you have the pic you can use Pythagoras's theorem.

\(\displaystyle \left\{\frac{x^{102}-1}{x^{3}-1}-\frac{x^{101}-x^{2}}{x^{3}-1}\right\}(x^{2}+x+1)=x^{101}+x+1, \\ \text{so} \\ A+B=2.\)

The two fractions inside the curly brackets are both the sums of GP's.

The first is the sum of

 \(1+x^{3}+x^{6}+ \dots +x^{99},\)

and the second the sum of

\(x^{2}+x^{5}+ \dots + x^{98}.\)

LMFAO!! The above is a descriptive case of mathematical fornication QED. I’m going to stroke from LMAO!

\(\lceil 2x \rceil\) may not be equal to \(2 \lceil x \rceil\). But indeed, this equation has no solutions.

(a) Because \(z\bar{z} = |z|^2\), \(z\bar{z} = 1^2\). This means \(\overline z= \dfrac1z\).

Similarly, with the exact same steps, \(\overline w= \dfrac1w\).

(b) We write z and w in polar form. Let \(z = \cos \theta + i\sin \theta\) and \(w = \cos \phi + i\sin\phi\).

\(\dfrac{z + w}{zw + 1} = \dfrac{(\cos \theta + \cos \phi) + i(\sin \theta + \sin \phi)}{(\cos \theta + i\sin \theta)(\cos \phi + i\sin \phi) + 1} = \dfrac{(\cos \theta + \cos \phi) + i(\sin \theta + \sin \phi)}{(\cos \theta \cos \phi - \sin \theta \sin \phi + 1) + i(\sin \theta \cos \phi + \cos \theta \sin \phi)}\)

Simplifying, \(\dfrac{z + w}{zw + 1} = \dfrac{((\cos \theta + \cos \phi) + i(\sin \theta + \sin \phi))(\cos(\theta + \phi) + 1 - i\sin(\theta + \phi))}{(\cos(\theta + \phi) + 1)^2+\sin^2 (\theta + \phi)} \)

This means \(\operatorname{Im}\left(\dfrac{z + w}{zw + 1}\right) = \dfrac{(\cos(\theta + \phi) + 1)(\sin \theta + \sin \phi) - (\cos \theta + \cos \phi)\sin(\theta + \phi)}{(\cos(\theta + \phi) + 1)^2 + \sin^2(\theta + \phi)}\)

Notice that 

\((\cos(\theta + \phi) + 1)(\sin \theta + \sin \phi) \\= \sin \theta + \sin \phi + \sin \theta \cos \theta \cos \phi -\sin^2 \theta \sin \phi + \sin \phi \cos \theta \cos \phi - \sin \theta \sin^2 \phi\\=\sin \theta \cos^2 \phi + \sin \phi \cos^2 \theta+\sin \theta \cos \theta \cos \phi + \sin \phi \cos \theta \cos \phi\)

Also, 

\((\cos \theta + \cos \phi)\sin(\theta + \phi)\\ = \cos \theta \sin \theta \cos \phi + \cos \theta \sin \phi \cos \theta+\cos \phi \sin \theta \cos \phi + \cos \phi \sin \phi \cos \theta\\ = \cos \theta \sin \theta \cos \phi + \cos^2 \theta \sin \phi+\cos^2 \phi \sin \theta + \cos \phi \sin \phi \cos \theta\)

This means \((\cos(\theta + \phi) + 1)(\sin \theta + \sin \phi) = (\cos \theta + \cos \phi)\sin(\theta + \phi)\).

Then it immediately follows that \(\operatorname{Im}\left(\dfrac{z + w}{zw + 1}\right) = 0\), which means \(\dfrac{z + w}{zw + 1}\) is a real number.

Lia ate  2/3  of the pizza which means she left  1/3 .

So her brother received  1/3  of  6/8  of a whole pizza.

That is,

the amount her brother received  =  1/3 * 6/8  of a whole pizza

\(\dfrac13\times\dfrac68\ =\ \dfrac{1\times6}{3\times8}\ =\ \dfrac{6}{24 }\ =\ \dfrac{1}{4}\)

So...

her brother received  1/4  of a whole pizza

Let's draw a line straight down from  P  to the  x-axis  and call that point of intersection  S.

Now notice that the area of △PQR is equal to the area of △PQS minus the area of △PRS

Now since  P  is  6  units above the x-axis,

PS  =  6

From the picture we can see that PR has a steeper slope than PQ,

and so we can say the slope of PQ is  1  and the slope of PR is 3

Since  PQ  has a slope of  1,

rise / run  =  1

PS / QS  =  1

6 / QS  =  1

6  =  QS

QS  =  6

Since PR  has a slope of  3 ,

rise / run  =  3

PS / RS  =  3

6 / RS  =  3

6  =  3 * RS

2  =  RS

RS  =  2

Now we can find the area of  △PQS and the area of △PRS

area of △PQS   =   (1/2) * base * height   =   (1/2) * QS * PS   =   (1/2) * 6 * 6   =   18

area of △PRS   =   (1/2) * base * height   =   (1/2) * RS * PS   =   (1/2) * 2 * 6   =   6

And so...

area of △PQR   =   area of △PQS  -  area of △PRS   =   18 - 6   =   12

(Note... all areas are in square units)

Alternatively...we can do

area of  △PQR   =   (1/2) * base * height   =   (1/2) * QR * PS   =   (1/2) * 4 * 6   =   12

We first focus on the first and last part of the equality.

(y - 2)/3 = 24/(y - 1)

(y - 2)(y - 1) = 72

y^2 - 3y + 2 - 72 = 0

y^2 - 3y - 70 = 0

(y - 10)(y + 7) = 0

y = 10 or y = -7

Case 1: y = 10

Then (10 - 2)/3 = (1 - 5x)/(10 - 4).

i.e., (1 - 5x)/6 = 8/3

(1 - 5x) = 16

x = -3

So (x, y) = (-3, 10) is a solution.

Case 2: y = -7

Then (-7 - 2)/3 = (1 - 5x)/(-7 - 4)

i.e., 1 - 5x = 33

x = -32/5

So (x, y) = (-32/5, -7) is another solution.

Perpendicular: Product of slopes = -1.

Let m be the slope of the line perpendicular to y = -2x + 5.

Slope of y = -2x + 5 is -2.

This means \(-2\cdot m = -1\), i.e., \(m = \dfrac12\).

The required line passes through (5, 5) and the slope is 1/2. We use the point-slope form of the equation of a straight line.

The equation of the required line is 

\(y - 5 = \dfrac12 \left(x - 5\right)\\ 2y - 10 = x - 5\\ \boxed{x - 2y + 5 = 0}\)

The velocity of the wave, the frequency, and the wavelength is related by the formula: \(v = f\lambda\).

We substitute v = (speed of light) = c, f = \(7.5\cdot 10^{14}\).

\(3\cdot 10^8 = 7.5\cdot10^{14} \cdot \lambda\\ \lambda = 4 \cdot 10^{-7}\text{ m}\)

Then, we use Einstein's equation \(E = hf\) to find the energy of the photon.

\(E = 6.6262\cdot 10^{-34}\cdot 7.5\cdot 10^{14}\\ E = 4.96965 \cdot 10^{-19} J\)

area of sector / area of circle  =  100 / 360

Let  r  be the radius of the circle   so   the area of the circle  =  π r²

Substitute  250π  in for the area of the sector and substitute  π r²  in for the area of the circle.

250π  /  (π r²)   =   100 / 360

                                               Cancel out the  π's  in the first fraction

250 / r²   =   100 / 360

                                               Flip both sides of the equation

r² / 250   =   360 / 100

                                               Multiply both sides of the equation by  250

r²   =   360 / 100 * 250

                                               Simplify the right side of the equation.

r²   =   900

                                               Take the positive square root of both sides.

r   =   30

What do I do with this function? 

Find the derivative? Find the domain? Please specify.

hello pls someone help me

f(x)=lg(2-7x)

Hello mongoloco!

Are you looking for the zero point ?!

\(f(x)=lg(2-7x)=0\)  |  de-logaritmize

\(2-7x=10^0\)                  |  \(\mathbb R^0\in \{1\}\) 

\(2-7x=1\)                      |  -2

\(-7x=1-2\)                   | \(\times (-1)\)

\(7x=1\)                              |  / 7

\(x=\dfrac{1}{7}\)

greeting

  !

Q1. 

Rearranging the equation,

\(x^2 - 7x - 30 = 0\)

Then, using the quadratic formula, 

\(x = \dfrac{7 \pm \sqrt{7^2 - 4\cdot 1 \cdot (-30)}}{2}\)

I will leave the simplification for you.

Q2. Similar to Q1. If you understand what I did in Q1, then it's not hard. Just repeat what I did in Q1.

Q3. Keyword: "the ... equation ... has a double root."

The discriminant of a quadratic equation which has a double root is 0.

\(b^2 - 4\cdot 2 \cdot 18 = 0\\ b^2 - 12^2 = 0\\ (b - 12)(b + 12) = 0\)

Then this means b - 12 = 0 or b + 12 = 0. This is just one step from the answer, I will let you figure out this part.

Q4. This is a root relation problem. We will use the formulae for relations of roots.

\(\begin{cases} r + s = -\dfrac{(-6)}{1} = 6\\ rs = \dfrac{2}1 = 2 \end{cases}\)

We then express (r - s)^2 in terms of r + s and rs. 

\((r - s)^2 = r^2 - 2rs + s^2 = (r^2 + 2rs + s^2) - 4rs = (r + s)^2 -4rs\)

You can substitute the values of r + s and rs directly.

Q5. Similar to Q1. Repeat what I did.

Q6. 

\(\dfrac c{c - 5} = \dfrac4{c - 4}\)

Cross-multiplying,

\(c(c - 4) = 4(c- 5)\)

Expanding,

\(c^2 - 4c = 4c - 20\)

Moving everything to the left-hand side,

\(c^2 - 8c + 20 = 0\)

Then repeat what I did in Q1.

P.S. The entire "Quadratic Equation" part of the syllabus is for solving equations of the form ax^2 + bx + c = 0, so make sure you put the question in this form before you proceed.

P.P.S. Some parts need knowledge of discriminant \(\Delta = b^2 - 4ac\) and the root relation formulae. 

P.P.P.S. If you don't understand any part of my solutions, feel free to ask me.

Let the length of the hypotenuse in inches  =  h

Then by the Pythagorean theorem,

5²  +  4.5²   =   h²

25 + 20.25   =   h²

45.25   =   h²

√[ 45.25 ]   =   h

The perimeter of a triangle is the sum of the length of all its sides. So...

perimeter in inches   =   5  +  4.5   +  √[ 45.25 ]

perimeter in inches   =   9.5  + √[ 45.25 ]

perimeter in inches   ≈   16.227

It's true though

Fallen AngelGirl, this electronic forum is neither a sacred scroll nor a roll of toilet paper,  so you do not need to scroll ten lines to protect your writing from any fingers and their prints that may come near the fringes.

Your space of ten lines does not protect the forum from your blatant plagiarism. 

I instantly recognized Melody’s work

A solid right prism has a height of 16 as shown. Also, its bases are equilateral triangles with a side length of 12. Points X Y Z   and are the midpoints of edges  AC, BC, DC respectively. A part of the prism above is sliced off with a straight cut through points  X Y Z. Determine the volume of solid CXYZ, the part that was sliced off.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

A solid CXYZ is a pyramid with an equilateral triangle as its base.

Solid height       h = 8

Base area          A = ¹/₄ * √3 * a²              a = 6              

Solid volume      V = ¹/₃ * A * h         

12 dollars sorry for being this late to the party

Thank you! I'll definitely check the link out. I could use some extra practice in this area.

Thanks textot  

Here is another method:

\(f(100)=100^2-100+2010\)

\(f(101)=101^2-101+2010\\ f(101)=(100+1)^2-101+2010\\ f(101)=100^2+200+1-101+2010\\ f(101)=100^2-100+200+2010\\ f(101)=f(100)+200\\\)

Now my problem has changed to, what is the highest common factor of

  \(100^2-100+2010 \qquad and \qquad 200\)

10 goes into both so I will divide by 10

\(1000-10+201 \qquad and \qquad 20\)

The last digit of the first one is 1.   Nothing ending in 1 will have any common factors with 20, (it would have to end in a multiple of 2 or 5)

So

The hight common factor is 10.

I just realized i made a typo, because x^2-x = x(x-1), but the answer remains the same

Multiply top and bottom of the given  by   3 sqrt5 - 2sqrt11   to get

2 (3sqrt5 - 2 sqrt11 ) / (45-44)

= (6 sqrt5 - 4 sqrt11)/1                     ....take it from here...